! of 1 35

!

! of 2 35

Bone

By

Ian Beardsley

! of 3 35

!

! of 4 35

Introduction"

In comparing biological life to artiﬁcial intelligence, I ﬁnd that the instance of bone is the most

compelling. It is at this point that I suggest biological life can be taken as a mathematical struc-

ture actually using physical aspects of biological life such as molar mass, density, and atomic

radii as the values of the variables. Life seems to only present itself this way if we compare it to

another construct, like artiﬁcial intelligence. The other extraordinary thing is that bone, which is

the fundamental framework around which life is built (muscles are attached to it, skin wrapped

around it, organs embedded in it) is described by the fundamental framework around which

mathematics is built, algebra. Thus, in this paper dedicated to bone, we have all the framework

of fundamental mathematics following from it (ratios, proportions, completing the square, qua-

dratic equations, the golden ratio). As we progress to the form built on the skeleton (muscle on

bone) we proceed to the next layer of mathematics, calculus, diﬀerential equations, and vector

calculus). For instance muscle action is like a damped harmonic oscillator in that the force on

the muscle in moving a load is proportional the distance it contracts, and the solution of such a

diﬀerential equation makes use of the fundamental framework of algebra, namely its solutions

are obtained by ﬁnding the exponents of e with algebra’s quadratic equation or, equivalently,

the factorization of a quadratic, or completing the square. More primary to bone are the amino

acids and biological elements such as C, N, O, H, which when compared to the AI elements Si,

Ge, P, B, As, Ga, of which aspects such as bone are built, and semiconductor components are

built for logic gates in AI, these ﬁnd their expression in an elegant set of equations I laid out in

earlier work (AI Biodesign, 2019) which make use of ratios and proportions, such as the golden

ratio (a/b=b/c, a=b+c) which are building blocks to bone’s algebra just as in they are building

blocks to things like amino acids, DNA (in the case of biological life) and Si, Ge, P, B, As, are

building blocks to circuit components in AI. The equations that follow on all layers in terms of

molar mass, density, and atomic radii, may be of such parallel construct in the need for func-

tion in that molar mass, density, and atomic radii, determine the properties of elements and

their ensuing compounds.!

! of 5 35

Part 1"

It would seem there is some possibility that as life goes from the fundamental framework to

form, so does the math that describes it. I make no attempt to understand why the composi-

tion of life is aesthetically pleasing and how that came about, but I think it is due to the need

for function. The structure herein found arose when I was comparing biological life to artiﬁcial

intelligence. The case for bone was so interesting to me that I decided to proceed to muscle

and skin. It is because of this structure only revealing itself when comparing the biological to

AI, that I might suggest you can only speak about what life is relative to another construct, like

AI. Especially where awareness is concerned if we consider the Turing test.!

! of 6 35

In my exploration of the connection between biological life and AI the most dynamic compo-

nent is that of bone. It aﬀords us the opportunity to look at:"

Multiplying Binomials"

Completing The Square"

The Quadratic Formula"

Ratios"

Proportions"

The Golden Ratio"

The Square Root of Two"

The Harmonic Mean!

! of 7 35

Density of silicon is Si=2.33 grams per cubic centimeter."

Density of germanium is Ge=5.323 grams per cubic centimeter."

Density of hydroxyapatite is HA=3.00 grams per cubic centimeter."

This is"

where "

Where HA is the mineral component of bone, Si is an AI semiconductor material and Ge is an

AI semiconductor material. This means"

"

The harmonic mean between Si and Ge is HA,…"

"

This is the sextic,…"

"

Which has a solution"

"

Where x=Si, and y=Ge. It can be solved with the online Wolfram Alpha computational engine.

But,…"

"

!

3

4

Si +

1

4

Ge ≈ H A

HA = Ca

5

(PO

4

)

3

OH

Si

HA

Si +

[

1 −

Si

HA

]

Ge = H A

2SiGe

Si + Ge

≈ H A

x

2

(x + y)

4

− x y(x + y)

4

+ 2x y

2

(x + y)

3

− 4x

2

y

2

(x + y)

2

= 0

Si

Ge

=

1

2 + 1

1

HA

2

Si

2

−

Ge

HA

2

Si +

[

Ge

HA

− 1

]

= 0

Si =

1

2

Ge

±

HA

Ge

HA

2

−

4Ge

HA

+ 4

Si = Ge − HA

! of 8 35

"

Si

HA

Si +

[

1 −

Si

HA

]

Ge = H A

Si

2

HA

+ Ge −

Si

HA

Ge ≈ H A

1

HA

Si

2

−

Ge

HA

Si + Ge ≈ H A

1

HA

2

Si

2

−

Ge

HA

2

Si +

Ge

HA

≈ 1

1

HA

2

Si

2

−

Ge

HA

2

Si +

Ge

HA

− 1 ≈ 0

1

HA

2

Si

2

−

Ge

HA

2

Si +

[

Ge

HA

− 1

]

= 0

! of 9 35

"

We see that the square of the binomial is a quadratic where the third term is the square of one

half the middle coeﬃcient. This gives us a method to solve quadratics called completing the

square:"

"

!

(x + a)(x + a) = x

2

+ 2a x + a

2

(x + a)

2

= x

2

+ 2a x + a

2

a x

2

+ bx + c = 0

a x

2

+ bx = − c

x

2

+

b

a

x = −

c

a

(

1

2

b

a

)

2

=

1

4

b

2

a

2

x

2

+

b

a

x +

1

4

b

2

a

2

= −

c

a

+

1

4

b

2

a

2

(

x +

1

2

b

a

)

2

=

b

2

− 4ac

4a

2

x +

b

2a

=

±

b

2

− 4ac

2a

x =

−b

±

b

2

− 4ac

2a

! of 10 35

"

!

1

HA

2

Si

2

−

Ge

HA

2

Si +

[

Ge

HA

− 1

]

= 0

x =

−b

±

b

2

− 4ac

2a

a =

a

HA

2

b = −

Ge

HA

2

c =

[

Ge

HA

− 1

]

b

2

− 4ac =

Ge

2

HA

4

− 4

1

HA

2

[

Ge

HA

− 1

]

=

Ge

2

HA

4

−

4Ge

HA

3

+

4

HA

2

=

1

HA

2

[

Ge

2

HA

2

−

4Ge

HA

+ 4

]

b

2

− 4ac =

1

HA

(

Ge

HA

− 2

)

2

x =

Ge

HA

2

±

1

HA

[

Ge

HA

− 2

]

2

HA

2

=

1

2

Ge

±

1

2

HA

[

Ge

HA

− 2

]

=

1

2

Ge

±

1

2

Ge − HA

Si =

1

2

Ge +

1

2

Ge − HA

Si = Ge − HA

! of 11 35

"

Si ≈ Ge − HA

HA ≈

2SiGe

Si + Ge

Si ≈ Ge −

2SiGe

Si + Ge

(Si + Ge)Ge

Si + Ge

−

(Si + Ge)Si

Si + Ge

−

2SiGe

Si + Ge

= 0

Ge

2

− 2SiGe − Si

2

Si + Ge

= 0

x

2

− 2x y − y

2

= 0

x

2

− 2x y = y

2

x

2

− 2x y + y

2

= 2y

2

(x − y)

2

= 2y

2

x − y =

±

2y

x = y + 2y

x = y(1 + 2)

x

y

= 1 + 2

y

x

=

1

2 + 1

Si

Ge

≈

1

2 + 1

! of 12 35

A ratio is and a proportion is which means a is to b as b is to c."

The Golden Ratio "

and. "

or "

"

a

b

a

b

=

b

c

(

Φ

)

a

b

=

b

c

a = b + c

ac = b

2

c =

b

2

a

a = b +

b

2

a

b

2

a

− a + b = 0

b

2

a

2

− 1 +

b

a

= 0

(

b

a

)

2

+

b

a

− 1 = 0

(

b

a

)

2

+

b

a

+

1

4

= 1 +

1

4

(

b

a

+

1

2

)

2

=

5

4

b

a

= −

1

2

±

5

2

b

a

=

5 − 1

2

a

b

=

5 + 1

2

ϕ =

5 − 1

2

Φ =

5 + 1

2

ϕ =

1

Φ

! of 13 35

The mineral component of bone hydroxyapatite (HA) is"

"

The organic component of bone is collagen which is"

"

We have"

"

%"

"

!

Ca

5

(PO

4

)

3

OH = 502.32

g

mol

C

57

H

91

N

19

O

16

= 1298.67

g

mol

Ca

5

(PO

4

)

3

OH

C

57

H

91

N

19

O

16

= 0.386795722

ϕ = 0.618033989

1 − ϕ = 0.381966011

Ca

5

(PO

4

)

3

OH

C

57

H

91

N

19

O

16

≈ (1 − ϕ)

0.381966011

0.386795722

100 = 98.75

Si

Ge

=

28.09

72.61

= 0.386861314 ≈ (1 − ϕ)

Si

Ge

≈

Ca

5

(PO

4

)

3

OH

C

57

H

91

N

19

O

16

! of 14 35

Part 2"

In comparing biological life to artiﬁcial intelligence I found the case for bone very compelling for

life as a mathematical construct. It is really quite interesting, if not aesthetically pleasing in its

structure. I covered that in my paper Life As A Mathematical Construct, here I revisit bone us-

ing those ﬁndings in terms of the intermembral index which compares the forelimbs of verte-

brates to their hindlimbs. A ratio greater than one means the forelimbs are longer than the

hindlimbs and less than one the hindlimbs are longer. It is this ratio that tells paleontologists a

great deal about the manner of propulsion of a vertebrate.!

! of 15 35

!

! of 16 35

We begin by looking at hydroxyapatite (HA) the mineral component of bone and comparing it to

silicon (Si) and germanium (Ge) the most basic components of artiﬁcial intelligence (AI)."

Density of silicon is Si=2.33 grams per cubic centimeter."

Density of germanium is Ge=5.323 grams per cubic centimeter."

Density of hydroxyapatite is HA=3.00 grams per cubic centimeter."

This is"

where "

This gives"

and. "

Which is the sextic"

"

Which has solution"

"

Because if y=Si, and x=Ge, we have"

"

3

4

Si +

1

4

Ge ≈ H A

HA = Ca

5

(PO

4

)

3

OH

Si

HA

Si +

[

1 −

Si

HA

]

Ge = H A

2SiGe

Si + Ge

≈ H A

x

2

(x + y)

4

− x y(x + y)

4

+ 2x y

2

(x + y)

3

− 4x

2

y

2

(x + y)

2

= 0

Si

Ge

=

1

2 + 1

Si ≈ Ge − HA

HA ≈

2SiGe

Si + Ge

Ge

2

− 2SiGe − Si

2

Si + Ge

= 0

x

2

− 2x y − y

2

= 0

y

x

=

1

2 + 1

! of 17 35

Thus we see that if i is the intermembral index and Si and Ge subscripted with rho are the den-

sities of silicon and germanium, respectively. Thus,…"

"

And,…"

"

Now we introduce the organic component of bone, collagen:"

"

We have"

"

%"

"

By molar mass. Where, is the golden ratio conjugate, is recurrent throughout ratios in

vertabrates determined by bone lengths.!

i ≈ 0.7

Si

ρ

Ge

ρ

≈

1

2 + 1

1

i

≈

1

2 + 1

+ 1

Ca

5

(PO

4

)

3

OH = 502.32

g

mol

C

57

H

91

N

19

O

16

= 1298.67

g

mol

Ca

5

(PO

4

)

3

OH

C

57

H

91

N

19

O

16

= 0.386795722

ϕ = 0.618033989

1 − ϕ = 0.381966011

Ca

5

(PO

4

)

3

OH

C

57

H

91

N

19

O

16

≈ (1 − ϕ)

0.381966011

0.386795722

100 = 98.75

Si

Ge

=

28.09

72.61

= 0.386861314 ≈ (1 − ϕ)

Si

Ge

≈

Ca

5

(PO

4

)

3

OH

C

57

H

91

N

19

O

16

ϕ

! of 18 35

Part 3"

I ﬁnd 1/4 coupled with 3/4 is recurrent throughout Nature and that where there is (1/4, 3/4)

there is (1/3, 2/3) where 1/3 = 1- 2/3. An example of this would be in Mendelian genetics. 3/4

of the oﬀ spring express a recessive trait. This is because while the male has XY sex chromo-

somes the female has XX sex chromosomes. Thus,…"

XY + XX = 3X + Y or 1Y: 3X=>1/3 and 1-1/3=2/3"

But we can say,…"

XY + XX = X + X + X + Y = 4 elements"

And,…"

"

And, 1-1/4=3/4."

This is how (1/4, 3/4) is coupled with (1/3, 2/3)!

Y

X + X + X + Y

=

1

4

elements

! of 19 35

Air is about 25% oxygen gas (O2) by volume and 75% nitrogen gas (N2) by volume meaning

the molar mass of air as a mixture is:"

"

By density we ﬁnd"

"

(Density of Ge is 5.323 g/cm^3 and of Si is 2.33 g/cm^3)"

is the mineral component of bone."

This is the source of our equations:"

"

The density of HA is about 3.00 g per cubic centimeter."

"

Thus if we consider"

"

Then we should consider sea water, or ocean:"

Thus we have,…"

0.25O

2

+ 0.75N

2

≈ air

0.25Ge + 0.75Si ≈ H A

HA = hydrox yapat ite = Ca

5

(PO

4

)

3

OH

Si

HA

(Si) +

[

1 −

Si

HA

]

(Ge) ≈ HA

2(Si)(Ge)

Si + Ge

≈ HA

Si

HA

(Si) +

[

1 −

Si

HA

]

(Ge) ≈ HA

1

HA

2

Si

2

−

Ge

HA

2

Si +

[

Ge

HA

− 1

]

= 0

Si =

1

2

Ge

±

HA

Ge

HA

2

−

4Ge

HA

+ 4

Si ≈ Ge − HA

1

4

O

2

+

3

4

N

2

≈ air

! of 20 35

"

The Ancient Greeks categorized Nature by 4 elements they called Earth, Air, Water, and Fire. In

a sense this was a way of saying solid, gas, liquid, energy. The Earth crust is highest in oxygen,

mainly because a lot of that is in water (hydrogen combined with oxygen). We have"

Oxygen (O) = 467,100 ppm"

Silicon (Si) = 276,900 ppm"

Aluminum (Al) = 80,700 ppm"

Iron (Fe) = 50,500 ppm"

If we say metal is Al +Fe = 131,200 and dirt is 276,900 then metal+dirt=408,100."

Dirt= 276,900/408,100 = 67.85%"

Metal = 131200/408,100 = 32%"

Thus,…"

0.67(dirt) + 0.33(metal) = earth or"

"

Is (1/3, 2/3) by parts."

"

Is (1/4, 3/4) by molar mass."

"

For Sea or ocean water by molar mass."

"

By density, where Ge and Si are mineral and sand components.!

NaCl = 58.44g/m olan d H

2

O = 18.02g/mol

1

3

NaCl ≈ H

2

Oan d

2

3

H

2

O ≈ Carbon(C )12.01g/mol

2

3

H

2

O + NaCl

NaCl

+

1

3

H

2

O

NaCl

≈ 1

1

3

metal +

2

3

dir t ≈ ear th

1

4

O

2

+

3

4

N

2

≈ air

2

3

H

2

O + NaCl

NaCl

+

1

3

H

2

O

NaCl

≈ 1

1

4

Ge +

3

4

Si = bone

! of 21 35

Why notice that the density of Si divided by the density of HA is about 3/4 and write"

"

And then use that with"

"

When it means we have to solve the sextic"

"

When we could have just noticed "

"

And then merely have to solve the quadratic"

"

The reason is that by ﬁrst looking at the ratios of our densities instead of their diﬀerences we

have bone as part of the Ancient Greek elements and that takes us to archaeology. What is Ar-

chaeology but the study of the decomposition of bone in its exposure to earth, air, and water?"

I found it interesting that bone was appearing in the context of Earth, Air, and Water. I knew

that the oldest skeletons of humans and their ancient ancestors were unearthed in Africa, and

that Chinese archaeologists wanted to ﬁnd skeletons in China just as old so they could show

they arose independently of any other people. The argument as to why they had not found any-

thing as old as in Africa was that Africa is drier so skeletons preserve longer. And, this brings

me to my point:"

The organic component of bone is collagen and decomposes early on. What is left is the min-

eral component (hydroxylapatite, HA). The amount of bone in eight weeks that decays in soil

(Earth) is the same as the amount of bone that decays in air in two weeks (CAP, 1986) and the

primary factors in bone decomposition are soil, air, and water. Mendelian genetics are are at

the core of biological evolution and the study of that is the study of bone decomposition. I

need only present now, my earlier work. The mineral component (HA) decomposes in exposure

to water into calcium and phosphates."

Thus in humid conditions bones can decompose in decades and in dry conditions can be pre-

served for thousands of years. Francesco Berna, Alan Mathews, and Stephen Weiner report

(2004):"

We measured the ionic activity products at “steady-state” conditions and we identify a recrys-

tallization window between pH 7.6 and 8.1, which deﬁnes the conditions under which bone

crystals dissolve and reprecipitate as a more insoluble form of carbonated hydroxyl apatite. As

these conditions are common in nature, most fossil bones will not maintain their original crystals

with time.!

Si

HA

Si +

[

1 −

Si

HA

]

Ge = H A

2SiGe

Si + Ge

≈ H A

x

2

(x + y)

4

− x y(x + y)

4

+ 2x y

2

(x + y)

3

− 4x

2

y

2

(x + y)

2

= 0

Si = Ge − H A

x

2

− 2x y − y

2

= 0

! of 22 35

Part 4"

We have said"

"

So,…"

Or…"

"

We have also said"

"

In order for Si and Ge to semiconductor they must be doped. Often the doping agents are

phosphorus (P) and boron (B). We ﬁnd by atomic radius where SI_R=110pm, Ge_R=125pm,

P_R=100pm and B_r=85pm"

or,…. "

1

i

≈

1

2 + 1

+ 1 ≈ 2

Si

ρ

Ge

ρ

≈

1

2 + 1

1

i

≈

Si

ρ

Ge

ρ

+ 1

1

i

≈

Si

ρ

+ Ge

ρ

Ge

ρ

i ≈

Ge

ρ

Si

ρ

+ Ge

ρ

Si

M

Ge

M

≈

Ca

5

(PO

4

)

3

OH

C

57

H

91

N

19

O

16

Ge

ρ

Si

ρ

+ Ge

ρ

≈

2

= sin

π

4

= cos

π

4

Si

M

Ge

M

≈ (1 − ϕ)

Si

M

≈ (1 − ϕ)Ge

M

Si

M

≈

Ca

5

(PO

4

)

3

OH

C

57

H

91

N

19

O

16

Ge

M

P

R

+ B

R

Si

R

≈ Φ

Si

R

P

R

+ B

R

≈ ϕ

! of 23 35

But if we take the arithmetic mean between the geometric mean of P_M and B_M and the har-

monic mean and divide by Si_M we ﬁnd that,…"

"

Which yields,…"

"

Thus,…"

"

!

P

M

B

M

(P

M

+ B

M

) + 2P

M

B

M

2(P

M

+ B

M

)Si

M

≈ ϕ

2Si

R

P

R

+ B

R

Si

M

≈ P

M

P

M

+

2P

M

B

M

P

M

+ B

M

2Si

R

P

R

+ B

R

≈

C

57

H

91

N

19

O

16

Ca

5

(PO

4

)

3

OH

[

P

M

B

M

+

2P

M

B

M

P

M

+ B

M

]

1

Ge

M

Ge

ρ

Si

ρ

+ Ge

ρ

≈

2

! of 24 35

Thus since we have"

"

And, since we have,…"

"

Then we can write as well"

"

CO2 (carbon dioxide) is what animal life exhales when it breathes, which is taken up by plant

life to make the oxygen gas (O2) that it breathes in a process called photosynthesis where

plants get energy from the sun to make the monomer used to make its food the carbohydrate

CH2O at the basis of making the more complex sugars at the bottom of the food chain. The

reaction is:"

"

2Si

R

P

R

+ B

R

≈

C

57

H

91

N

19

O

16

Ca

5

(PO

4

)

3

OH

[

P

M

B

M

+

2P

M

B

M

P

M

+ B

M

]

1

Ge

M

Ge

ρ

Si

ρ

+ Ge

ρ

≈

2

CO

2

= 12.01 + 32.00 = 44.01

O

2

= 32.00

CO

2

O

2

=

44.01

32.00

= 1.375 ≈ 1.414 = 2

2Si

R

P

R

+ B

R

≈

C

57

H

91

N

19

O

16

Ca

5

(PO

4

)

3

OH

[

P

M

B

M

+

2P

M

B

M

P

M

+ B

M

]

1

Ge

M

2Ge

ρ

Si

ρ

+ Ge

ρ

≈

CO

2

O

2

CO

2

+ 2H

2

O + photons ⟶ CH

2

O + O

2

+ H

2

O

! of 25 35

Part 5

We look at the fourth Greek element, Fire. We choose peat as our tinder as it is naturally occur-

ring and would have been available to ancient humans for making ﬁre. It has one of the lowest

auto ignition temperatures, which is"

"

Peat has a speciﬁc heat of"

Or,… "

"

If we are to say building artiﬁcial intelligence (AI) is taking us back to our beginnings of making

ﬁre,… then"

If we are to have series of switches that are either on or oﬀ so we can encode in binary… we

need doped silicon or doped germanium (semiconductors like diodes). These have forward bi-

ases (current required to turn them on) of"

0.6 Volts for silicon and,"

0.3 Volts for germanium"

The work done in accelerating an electron through a potential of one volt is"

"

Then there are"

"

To turn on a silicon diode. For germanium it is"

"

These are"

227 ∘

C

1.88k J/kg ∘

K

0.45kcal /kg ∘

C

1kcal = 4148J

0.45kcal

kg ∘

C

4184J

kcal

kg

1000g

= 0.0018828J/g ∘

C

0.0018828J

g ∘

C

1 × 10

7

ergs

J

= 18828ergs/g ∘

C

g ∘

C

0.0018828ergs

1

227 ∘

C

= 0.012(gramsof peat)/erg

1.6 × 10

−19

Joules

(0.6V )(1.6 × 10

−19

J ) = 9.6 × 10

−20

J

(0.3J )(1.6 × 10

−19

) = 4.8 × 10

−20

J

! of 26 35

"

The density of dry peat is"

"

Thus we have the energy to turn on a silicon diode will burn"

"

Of peat. If we consider a strand of peat to be approximately cylindrical and to have a length of

12 times its radius then,…"

"

And,…"

"

Thus we say the energy to turn on a silicon diode in an AI circuit is about the energy to burn a

ten thousandth of a centimeter of peat, which from experience I think is the amount of tinder

we ignite with sparks from hitting together two pieces of ﬂint. Once that small piece is ignited,

it ignites the whole piece of tinder and that in turn ignites a pile of tinder used to set ﬁre to kin-

dling."

Thus in looking at the fourth Ancient Greek element ﬁre we return to our original problem of AI.!

(9.6 × 10

−20

J )

1

(1 × 10

7

ergs)

Joule

= 9.6 × 10

−13

ergs

(4.8 × 10

−20

J )

1

(1 × 10

7

ergs)

Joule

= 4.8 × 10

−13

ergs

(0.012gra msof peat)

erg

(9.6 × 10

−13

ergs)

1

= 1.152 × 10

−14

g

0.4g/cm

3

1.152 × 10

−14

g

1

cm

3

0.4g

= 2.88 × 10

−14

cm

3

V = πr

2

h

2.88 × 10

−14

cm

3

= π

(

1

12

h

)

2

h = (0.0218)h

3

h = 0.0001cm

! of 27 35

Part 6!

! of 28 35

!

! of 29 35

!

! of 30 35

Thus we have the very interesting equations

!

[

1 −

Si

ρ

Ge

ρ

]

≈

(1 − ϕ)

i

i = cos

π

4

, ϕ = cos

π

5

Si

ρ

Ge

ρ

Si

M

Ge

M

≈

(1 − ϕ)

(2i + 1)

! of 31 35

Journal Notes!

! of 32 35

!

! of 33 35

!

! of 34 35

!

! of 35 35

The Author